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As Far from Land as Possible in C++
Suppose we have one N x N grid containing only values like 0 and 1, where 0 represents water and 1 represents the land, we have to find a water cell such that its distance to the nearest land cell is maximized and return the distance. Here we will use the Manhattan distance − the distance between two cells (x0, y0) and (x1, y1) is |x0 - x1| + |y0 - y1|. If no land or water is present in the grid, then return -1.
| 1 | 0 | 1 |
| 0 | 0 | 0 |
| 1 | 0 | 1 |
Then the output will be 2, as the cell (1,1) is as far as possible from all the land with distance 2.
To solve this, we will follow these steps −
dir := [(1, 0), (-1, 0), (1, -1), (1, 1), (-1, 1), (-1, -1), (0, 1), (0, -1)]
dir2 := [(1, 0), (-1, 0), (0, 1), (0, -1)]
Define a map m. Define a queue q. n := row count and c := column count
for i in range 0 to n – 1
for j in range 0 to n – 1
if grid[i, j] is 1, then insert a pair (i, j) into q and put m[(i, j)] := (j ,i)
ret := -1
while the q is not empty
sz := size of q
while sz is not 0
temp := first element of q, delete first element from q
for k in range 0 to 3 −
nx := first value of temp + dir2[k, 0]
ny := second value of temp + dir2[k, 1]
if nx and ny are not in range of grid, or grid[nx, ny] is 1, then skip to the next iteration.
m[(nx, ny)] := m[temp]
ret := max of (distance of (nx, ny) and m(temp)) and ret
insert (nx,ny) into q
set grid[nx, ny] := 1
decrease sz by 1
return ret
Example(C++)
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
int dir[8][2] = {
{1, 0}, {-1, 0}, {1, -1}, {1, 1},
{-1, 1}, {-1, -1}, {0, 1}, {0, -1}
};
int dir2[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
class Solution {
public:
int calcDist(int x1, int y1, int x2, int y2){
return abs(x1 - x2) + abs(y1 - y2);
}
int maxDistance(vector<vector<int>>& grid) {
map < pair <int, int>, pair <int, int> > m;
queue < pair <int, int> > q;
int n = grid.size();
int c = n? grid[0].size() : 0;
for(int i = 0; i < n; i++){
for(int j = 0; j < c; j++){
if(grid[i][j] == 1){
q.push({i, j});
m[{i, j}] = {i, j};
}
}
}
int ret = -1;
while(!q.empty()){
int sz = q.size();
while(sz--){
pair <int, int> temp = q.front();
q.pop();
for(int k = 0; k < 4; k++){
int nx = temp.first + dir2[k][0];
int ny = temp.second + dir2[k][1];
if(nx < 0 || ny < 0 || nx >= n || ny >= c || grid[nx][ny]) continue;
m[{nx, ny}] = m[temp];
ret = max(calcDist(nx, ny, m[temp].first,
m[temp].second), ret);
q.push({nx, ny});
grid[nx][ny] = 1;
}
}
}
return ret;
}
};
main(){
vector<vector<int>> v1 = {{1,0,1},{0,0,0},{1,0,1}};
Solution ob;
cout << (ob.maxDistance(v1));
}Input
["alice,20,800,mtv","bob,50,1200,mtv"]
Output
2
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